In the figure, a very large plane sheet of positive charge is shown. $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are two points at distance $l$ and $2 l$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ at $P_{1}$ and $P_{2}$ respectively are :

(C) $\mathrm{E}_{1}=\mathrm{E}_{2}=\sigma / 2 \varepsilon_{0}$

(A) $\mathrm{E}_{1}=\sigma / \varepsilon_{0}, \mathrm{E}_{2}=\sigma / 2 \varepsilon_{0}$

(B) $\mathrm{E}_{1}=2 \sigma / \varepsilon_{0}, \mathrm{E}_{2}=\sigma / \varepsilon_{0}$

(D) $\mathrm{E}_{1}=\mathrm{E}_{2}=\sigma / \varepsilon_{0}$

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Electrostatics

In the figure, a very large plane sheet of positive charge is shown.

P_{1}

and

P_{2}

are two points at distance

l

and

2 l

from the charge distribution. If

σ

is the surface charge density, then the magnitude of electric fields

E_{1}

and

E_{2}

P_{1}

and

P_{2}

respectively are :

Answer: (C)

For an infinite plane sheet of charge with surface charge density

σ

, the electric field at any point is uniform and independent of the distance from the sheet:

E = \frac{σ}{2 ε _{0}}

This result comes from Gauss's law applied to a cylindrical Gaussian surface. Since the sheet is very large (effectively infinite), both

E_{1}

and

E_{2}

have the same magnitude regardless of their distances from the sheet.

Therefore:

E_{1} = E_{2} = \frac{σ}{2 ε _{0}}

Answer: (C)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option